A problem on related rates

Question

Calculusman is demonstrating his strength by squeezing a rod of solid iron. Throughout the process, the radius is decreasing at the rate of 1 cm/s, but the rod always remains a cylinder, with constant volume. Find the rate at which the height is increasing in cm/s, at the moment the rod has a radius of 4 cm and a height of 10 cm.

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I know that $\frac{dV}{dt} = 0 \frac{\text{cm}^3}{\text{s}}$, and $\frac{dr}{dt} = -1 \frac{\text{cm}}{s}$, but I don't know how I can use these definitions to my advantage. Help would be much appreciated!

Answer 1

Note that $$V = \pi r^2 h.$$ We now take the derivative of both sides to get $$V' = 2\pi r r' h + \pi r^2 h'.$$ As you noted $V' = 0$, $r' = -1$, and we are given that $r = 4$ and $h = 10$. Thus, putting our work together, we have that $$0 = 2\pi \cdot 4 \cdot (-1) \cdot 10 + \pi 16 h' \Rightarrow h' = 5 ~ \frac{\text{cm}}{\text{sec}}.$$

Answer 2

$$ \pi r^2 h = V= \text{constant} $$

Differentiate chain rule or product rule and simplify

$$\dfrac{r^2}{h} =\dfrac{-2rr'}{h'}$$

$$h'=\dfrac{-2r'h}{r}=\dfrac{-2\cdot -1 \cdot 10}{4} = 5 \text{cm/s}$$